, 2016

Solution 1. Suppose all positive divisors of $n$ can be arranged into a rectangular table of size $k \times l$ where the number of rows $k$ does not exceed the number of columns $l$. Let the sum of numbers in each column be $s$. Since $n$ belongs to one of the columns, we have $s \geqslant n$, where equality holds only when $n=1$.
For $j=1,2, \ldots, l$, let $d_{j}$ be the largest number in the $j$-th column. Without loss of generality, assume $d_{1}>d_{2}>\cdots>d_{l}$. Since these are divisors of $n$, we have
$$d_{l} \leqslant \frac{n}{l} \tag{1}$$
As $d_{l}$ is the maximum entry of the $l$-th column, we must have
$$d_{l} \geqslant \frac{s}{k} \geqslant \frac{n}{k} \tag{2}$$
The relations (1) and (2) combine to give $\frac{n}{l} \geqslant \frac{n}{k}$, that is, $k \geqslant l$. Together with $k \leqslant l$, we conclude that $k=l$. Then all inequalities in (1) and (2) are equalities. In particular, $s=n$ and so $n=1$, in which case the conditions are clearly satisfied.

Solution 2. Clearly $n=1$ works. Then we assume $n>1$ and let its prime factorization be $n=p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{t}^{r_{t}}$. Suppose the table has $k$ rows and $l$ columns with $1<k \leqslant l$. Note that $k l$ is the number of positive divisors of $n$ and the sum of all entries is the sum of positive divisors of $n$, which we denote by $\sigma(n)$. Consider the column containing $n$. Since the column sum is $\frac{\sigma(n)}{l}$, we must have $\frac{\sigma(n)}{l}>n$. Therefore, we have
$$\begin{aligned} \left(r_{1}+1\right)\left(r_{2}+1\right) \cdots\left(r_{t}+1\right) & =k l \leqslant l^{2}<\left(\frac{\sigma(n)}{n}\right)^{2} \\ & =\left(1+\frac{1}{p_{1}}+\cdots+\frac{1}{p_{1}^{r_{1}}}\right)^{2} \cdots\left(1+\frac{1}{p_{t}}+\cdots+\frac{1}{p_{t}^{r_{t}}}\right)^{2} . \end{aligned}$$
This can be rewritten as
$$\begin{equation*} f\left(p_{1}, r_{1}\right) f\left(p_{2}, r_{2}\right) \cdots f\left(p_{t}, r_{t}\right)<1 \tag{3} \end{equation*}$$
where
$$f(p, r)=\frac{r+1}{\left(1+\frac{1}{p}+\cdots+\frac{1}{p^{r}}\right)^{2}}=\frac{(r+1)\left(1-\frac{1}{p}\right)^{2}}{\left(1-\frac{1}{p^{r+1}}\right)^{2}}$$
Direct computation yields
$$f(2,1)=\frac{8}{9}, \quad f(2,2)=\frac{48}{49}, \quad f(3,1)=\frac{9}{8}$$
Also, we find that
$$\begin{aligned} & f(2, r) \geqslant\left(1-\frac{1}{2^{r+1}}\right)^{-2}>1 \quad \text{ for } r \geqslant 3 \\ & f(3, r) \geqslant \frac{4}{3}\left(1-\frac{1}{3^{r+1}}\right)^{-2}>\frac{4}{3}>\frac{9}{8} \quad \text{ for } r \geqslant 2, \text{ and } \\ & f(p, r) \geqslant \frac{32}{25}\left(1-\frac{1}{p^{r+1}}\right)^{-2}>\frac{32}{25}>\frac{9}{8} \quad \text{ for } p \geqslant 5 \end{aligned}$$
From these values and bounds, it is clear that (3) holds only when $n=2$ or 4 . In both cases, it is easy to see that the conditions are not satisfied. Hence, the only possible $n$ is 1 .